consider the quadratic function (f(x)=\frac{1}{5}x^{2}-5x + 12). which statements are true about the…

Explanation:

Step1: Calculate f(-10)

Substitute x = - 10 into $f(x)=\frac{1}{5}x^{2}-5x + 12$. $f(-10)=\frac{1}{5}\times(-10)^{2}-5\times(-10)+12=\frac{1}{5}\times100 + 50+12=20 + 50+12=82$.

Step2: Determine the shape of the graph

The general form of a quadratic function is $y = ax^{2}+bx + c$, and its graph is a parabola. Here $a=\frac{1}{5},b=-5,c = 12$, so the graph of $f(x)$ is a parabola.

Step3: Determine the opening - direction of the graph

For a quadratic function $y = ax^{2}+bx + c$, if $a>0$, the parabola opens up; if $a<0$, the parabola opens down. Since $a=\frac{1}{5}>0$, the graph of the function opens up.

Step4: Check if the point (20, - 8) is on the graph

Substitute x = 20 into $f(x)$: $f(20)=\frac{1}{5}\times20^{2}-5\times20 + 12=\frac{1}{5}\times400-100 + 12=80-100 + 12=-8$. So the graph contains the point (20,-8).

Step5: Check if the point (0,0) is on the graph

Substitute x = 0 into $f(x)$: $f(0)=\frac{1}{5}\times0^{2}-5\times0 + 12=12\neq0$. So the graph does not contain the point (0,0).

Answer:

The value of $f(-10)=82$, The graph of the function is a parabola, The graph contains the point $(20,-8)$.